OK, I've got it. There is no such local criterion for squareness.
Let be a field of characteristic not . Take the ring of triples , subject to the conditions that , and . Consider the element . If this were a square, its square root would have to be . But two of those 's would be the same sign, and evaluated at and at are not equal.
Now, to check that is everywhere locally a square. Geometrically, we are talking about three lines glued into a triangle. Any prime ideal has a neighborhood which is contained in the union of two neighboring lines, say the first two. On the first two lines, is a square root of .
For the suspicious, an algebraic proof. Set and let and be the cyclic permutations thereof. We have so, in any local ring, one of the must be a unit. WLOG, suppose that is a unit. Notice that . So, in a local ring where is a unit, is a square root of .
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